Let $G$ be a finite group, $p$ the smallest prime divisor of $|G|$, and $x\in G$ an element of order $p$. Suppose $h\in G$ is such that $hxh^{−1}=x^{10}$. Show that $p=3$.
I am trying to solve this problem using group actions. Let $H$ and $X$ be the subgroups of $G$ generated by the elements $h$ and $x$, respectively. We have a homomorphism $\varphi\colon H\to\mathrm{Sym}(X)$, where $\mathrm{Sym}(X)$ is the symmetric group of $X$. What group action should I take? Does the given equation in some way imply that we can take the conjugation action of $H$ on $X$ or vice versa? I am stuck here. This question has already been asked before but the answers use methods other than group actions. I am more inclined towards using group actions.
 A: $H$ acts on $X$ by conjugation. This follows from $hxh^{-1} \in X$: it's easy to check that this implies $h^ix^jh^{-i} \in X$ for all $i, j$, i.e. $h'x'h'^{-1} \in X$ for all $h' \in H$, $x' \in X$. Checking that this action satisfies the other definitions of an axiom is also easy to check (it's the same as checking that any conjugation action is a valid action).
So by orbit-stabiliser, $|\text{Orb}(x)|$ must divide $|H|$, and so must divide $|G|$. So, either $|\text{Orb}(x)| = |X| = p$, or $|\text{Orb}(x)| = 1$, since if $1 < |\text{Orb}(x)| < p$ then any prime factor of $|\text{Orb}(x)|$ will be a smaller prime divisor of $|G|$ than $p$.
But we cannot have $|\text{Orb}(x)| = |X| = p$, since then $\text{Orb}(x) = X$ - but the identity $e$ is not in the orbit of $X$, since $h'eh'^{-1} = e$ for all $h' \in H$. So we must have $|\text{Orb}(x)| = 1$, i.e. $h'xh'^{-1} = x$ for all $h' \in H$. In particular, $x^{10} = hxh^{-1} = x$, and so $x$ has order dividing $9$. Since $x$ has prime order, it must have order $3$, i.e. $p = 3$.
